JEE PYQ: Atomic Physics Question 48
Question 48 - 2019 (09 Apr Shift 1)
Taking the wavelength of first Balmer line in hydrogen spectrum ($n = 3$ to $n = 2$) as 660 nm, the wavelength of the 2nd Balmer line ($n = 4$ to $n = 2$) will be:
(1) 889.2 nm
(2) 488.9 nm
(3) 642.7 nm
(4) 388.9 nm
Show Answer
Answer: (2)
Solution
$\frac{1}{\lambda_1} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = \frac{5R}{36}$. $\frac{1}{\lambda_2} = R\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = \frac{3R}{16}$. $\frac{\lambda_2}{\lambda_1} = \frac{80}{108}$. $\lambda_2 = \frac{80}{108} \times 660 = 488.9$ nm.
BETA
